Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;
public class Main{
	public static void main(String args[]){
    	Scanner sc=new Scanner(System.in);
      	int n=sc.nextInt();
      	double ans=0;
      	double arr[]=new double[n+1];
      	for(int i=1;i<=n;i++) arr[i]=sc.nextDouble();
      	double grid[][]=new double[n+1][n+1];
      	for(int i=0;i<=n;i++){
        	for(int j=0;j<=n;j++){
            	if(i == 0 && j== 0) grid[i][j]=1;
              	else if(i == 0) grid[i][j]=0;
              	else if ( j == 0) grid[i][j]=grid[i-1][j]*(1-arr[i]);
              	else {
                	grid[i][j]=grid[i][j]+((1-arr[i])*grid[i-1][j] + (arr[i]*grid[i-1][j-1]));
                }
            }
        }
      	int start=n/2+1;
      	for(int i=start;i<=n;i++) ans+=grid[n][i];
      	System.out.println(ans);
          
    }
}